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User blog:Mh314159/A new notation for large numbers
n|11 = n|1 = n + 1 1|1x = 1|x = A(x) = Ack(x,x) n|mx = n|(n|m-1x) n|x = (n-1)|n|(x-1)x 2|2 = 1|32 = A(A(A2) = A(A(7)) 2|3 = 1|2|23 = AA(A(7))3. This is much greater than Graham's number and also greater than Friedmann's n(4). 2|4 = 1|2|34 = the number you get when you recursively apply the A function 2|3 times. This is greater than 3>3>3>3 in Conway right arrow notation. 3|2 = 2|42 = 2|2|2|2|2 (right associative) = 2|2|2|(A(A(7)) where 2|(A(A(7)) = a functional recursion pyramid of 1| functions (A(A(7)) levels high with 2|1 at the top. Second level from top = 2|2 = 1|32 = A(A(7)). Third level from top = 2|3 (see above). Fourth level from top = 2|4 (see above). Each level involves applying the 2| function a number of times equal to the value of the previous level. After determining the number generated at the bottom of (A(A(7)) levels, this becomes the argument M in 2|M and when this is evaluated as some number P, the value of 3|2 is 2|P 9|9= 8|9|89 = a recursion pyramid 9 layers high with a left foot of 8 constant except for the top layer where it is 9, and a right foot of 9 counting down to 1. The top two layers: 8|9|12 = 8|102 = 8|8|8|8|8|8|8|8|8|8|2 right associated. Since the last 8|2 = 7|92 = 7|7|7|7|7|7|7|7|7|2 we have 8|8|8|8|8|8|8|8|8|7|7|7|7|7|7|7|7|7|2 etc. And when we have worked this down to 1| functions and evaluated right to left we will have evaluated the second layer from the top. tilde indicates left associate ~a|b|c... is left associative, meaning (a|b)|c)... a|1b = a|b a|c1 = a+1 if c > 1, |c indicates c consecutive bars 1|c a = ~a|c-1 a|c-1 ... a with a|c-1 a instances of a, left associative 1||2 = ~2|2|...2 with 2|2 instances of 2 1|||a = ~a||a||...a etc. a|cb = (a-1)|c a|c (b-1) b a|c mb = a|ca|c ...a|cb with m instances of a, right associative 1||3 = ~3|3|...3 with 3|3 instances of 3 2||2 = 1||2||12 = 1||(1||(1||2)) = 1||(1||(2|2)) = 1||(~a|a|...a) with a = 2|2 occurring a|a times. 2||3 = 1||2||23 = 1||(1||(...1||3)) with 2||2 instances 4||2 = 3||4||12 = 3||52 = 3||(3||(3||(3||(3||2)))) 4||4 = 3||4||34 1|||4 = ~4||4...4 with 4||4 instances of 4 3|||4 = 2|||3|||34 etc. Sequential symbols {n} represents the nth symbol in a sequence of symbols (other notations can be used such as a defined sequence of letters). {1} = | single bar 1{n}a = a{n-1}a a a{n}1 = a+1 a{n}b = (a-1){n}a{n}(b-1)b superscripts and subscripts affect {n} exactly as they affect the bar function Using # as the second symbol ordinal: 2#2 = 1#32 = 1#(1#(1#2) using {2} as the second symbol ordinal and therefore representing the same number: 2{2}2 = 1{2}32 = 1{2}(1{2}(1{2}2) 2{3}2 = 1{3}2{3}12 = 1{3}32 = 1{3}(1{3}(1{3}2)) where 1{3}2 = 2{2}2 2 = 1{2}2 32 = 1{2}2(1{2}2 (1{2}2 2)) and 1{2}2 2 = ~2{2}2{2}...2 with 2{2}2 instances of 2, left associative. Notice that symbol ordinals only decrease when the number preceding them decrements to 1 and there are no subscripts nested symbol ordinals a b = a{...{n{n}n}...}b with a{n}b sets of braces a }b = a n}}...}}b with a b sets of braces a{n:1}b = a b with a{n}b braces surrounding n a{n:m}b = a b with a{a{n}b,m-1}b braces surrounding n a{n:m:p}b = a b with a{a{n:m}b,a{n:m}b,p-1}b braces surrounding n etc. Category:Blog posts